//172.阶乘后的零
//https://leetcode.cn/problems/factorial-trailing-zeroes
class Solution {
public:
    int trailingZeroes(int n) {
        //2*5=10 求2和5个数的最小值
        //5的个数不会超过2 所以选择求5的个数
        int ret = 0;
        while(n)
        {
            //求当前n中5的个数 再求下一波
            n /= 5;
            ret += n;
        }
        return ret;
    }
};